What is the integral of (4x^{1/3}+6x^{-7}+10) ?

The integral of (4x^{1/3}+6x^{-7}+10) is 3x^{4/3}-1/(x^6)+10x+C

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integral of (4x^{1/3}+6x^{-7}+10)

\int (4 x^{\frac{1}{3}} + 6 x^{- 7} + 10) d x

3 x^{\frac{4}{3}} - \frac{1}{x ^{6}} + 10 x + C

Solution steps

\int 4 x^{\frac{1}{3}} + 6 x^{- 7} + 10 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4 x^{\frac{1}{3}} d x + \int 6 x^{- 7} d x + \int 10 d x

\int 4 x^{\frac{1}{3}} d x = 3 x^{\frac{4}{3}}

\int 6 x^{- 7} d x = - \frac{1}{x ^{6}}

\int 10 d x = 10 x

= 3 x^{\frac{4}{3}} - \frac{1}{x ^{6}} + 10 x

Add a constant to the solution

= 3 x^{\frac{4}{3}} - \frac{1}{x ^{6}} + 10 x + C

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What is the integral of (4x^{1/3}+6x^{-7}+10) ?
The integral of (4x^{1/3}+6x^{-7}+10) is 3x^{4/3}-1/(x^6)+10x+C